A Saturn year is 29.5 times the earth year.

Solution:

Distance of the Earth from the Sun, $r_{\mathrm{e}}=1.5 \times 10^{8} \mathrm{~km}=1.5 \times 10^{11} \mathrm{~m}$'

Time period of the Earth $=T_{e}$

Time period of Saturn, $T_{S}=29.5 T_{e}$

Distance of Saturn from the Sun $=r_{\mathrm{s}}$

From Kepler's third law of planetary motion, we have

$T=\left(\frac{4 \pi^{2} r^{3}}{\mathrm{G} M}\right)^{\frac{1}{2}}$

For Saturn and Sun, we can write

$\frac{r_{s}^{3}}{r_{e}^{3}}=\frac{T_{s}^{2}}{T_{e}^{2}}$

$r_{s}=r_{e}\left(\frac{T_{s}}{T_{e}}\right)^{\frac{2}{3}}$

$=1.5 \times 10^{11}\left(\frac{29.5 T_{e}}{T_{e}}\right)^{\frac{2}{3}}$

$=1.5 \times 10^{11}(29.5)^{\frac{2}{3}}$

$=1.5 \times 10^{11} \times 9.55$

$=14.32 \times 10^{11} \mathrm{~m}$

Hence, the distance between Saturn and the Sun is $1.43 \times 10^{12} \mathrm{~m}$.